The Second Incompleteness Theorem says that if $T$ is a consistent (computably) axiomatizable theory which extends IΣ1, then $\mathrm{Con}(T)$ is not provable from $T$. By analogy with computability theory, the stronger theory $T + \mathrm{Con}(T)$ can be thought of as the "jump" of $T$. To abuse this analogy, I will use $T'$ to denote the theory $T + \mathrm{Con}(T)$. I will write $T \leq S$ when $S$ proves every axiom of $T$; I will also write $S \equiv T$ (resp. $T < S$) when $T \leq S$ and $S \leq T$ (resp. $S \nleq T$).

It is well-known that if $T$ is consistent there are plenty of axiomatizable theories $S$ such that $T < S < T'$. In the following questions $H$ will denote an operator (like $\mathrm{Con}$) that uses the computable axiomatization of $T$ to produce a sentence $H(T)$. I will write $T^H$ for the theory $T + H(T)$.

Is there a computable operator $H(T)$ such that $T < T^H < T'$ for every consistent axiomatizable theory $T$ extending IΣ1? Is there such an operator which moreover satisfies that $T \equiv S$ implies $T^H \equiv S^H$?

Is there a computable operator $H(T)$ such that $(T^H)^H \equiv T'$ for every consistent axiomatizable $T$ extending IΣ1? Is there such an operator which moreover satisfies that $T \equiv S$ implies $T^H \equiv S^H$?

Question 1 asks for a uniform solution to the analogue of Post's Problem for axiomatizable theories. Question 2 asks for a uniform "half-jump" operator.

A little-known fact: Q is not strong enough for the second incompleteness theorem as it is usually proved, because Q doesn't prove the Hilbert-Bernays conditions on the Bew predicate. In fact, I read recently that Q does not verify that Bew is closed under modus ponens. I have not read the proof, though. Sigma^0_1 induction is apparently enough for the theory to verify the Hilbert-Bernays conditions, but not Sigma^0_0 induction. I only learned this when I had to actually document the hypotheses required for the incompleteness theorems this winter to help out an undergrad.
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Carl MummertMay 30 '10 at 3:01

Thanks Carl! I just replaced Q by ISigma_1 as you suggested.
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François G. Dorais♦May 30 '10 at 3:20

François, could you clarify what sense of computability you want here? Shall we assume that T is given by finitely many axioms over the base theory? Or do you want us to work with a program enumerating T?
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Joel David HamkinsMay 30 '10 at 20:53

Joel, I don't think it matters much what type of machine enumeration you use, but if you find it more comfortable you can assume that all enumerations are primitive recursive. Any computable enumeration is equivalent to a primitive recursive one via padding tricks.
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François G. Dorais♦May 30 '10 at 21:02

2 Answers
2

The answer to #1 is basically yes, because the proof that the Lindenbaum algebra above T is atomless is completely constructive.

Start with a (consistent) theory T to which the second incompleteness theorem applies, which means that T + ~Con(T) is also consistent. Then there is a sentence S such that T + ~Con(T) neither proves nor disproves S (using the first incompleteness theorem via Rosser's trick). So T + ~Con(T)$\land$~S is stronger than T + ~Con(T), but is still consistent. This means that T + ~(Con(T)$\lor$S) is consistent, so T + Con(T)$\lor$S is stonger than T.

The premise in your question that the Con operator itself
has the desired property and serves as a jump operator is
not universally true among the theories you consider.
Specifically, you seem to assume that because
$\text{Con}(T)$ is not provable in $T$, that
$T+\text{Con(T)}$ is consistent. But this is not correct,
because perhaps $T$ actually proves $\neg\text{Con}(T)$.
One easy instance of this is the theory
$T=PA+\neg\text{Con}(PA)$, which is consistent by the 2nd
Incompleteness Theorem, but clearly proves
$\neg\text{Con}(PA)$ and hence also $\neg\text{Con}(T)$.
Thus, as weird as it sounds, $T$ is a consistent theory
that proves its own inconsistency. In this case your theory
$T'$ is inconsistent and the jump failed. Carl's theory
$T^H$ in this case is consistent, but upon inspection you
will find that it is equivalent to $T$. So for this theory
$T$, your theory $T'$ jumped into inconsistency, and his
theory didn't jump at all.

One can similarly replace $PA$ here with any representable
theory $T_0$ and arrive at similar counterexamples, densely above any theory.

You can fix the question by
considering only the case where $T'$ is consistent, which
is surely what you had in mind. In this event, you would
only apply the jump when it happens to arrive at a
consistent theory. Since this question is not decidable
from a presentation of the theory, however, even from a
finite axiomatization, it may affect your motivation for
considering computable versions of the half-jump, since even the
full jump is not computable.

For this reason, and also because there is something a
little arbitrary about having the jump only partially
defined, it may be that a more robust jump arises from the
Rosser sentence---there is no proof of me without a
shorter proof of my negation---instead of $\text{Con}(T)$?
This would put you back into the universal domain of all
representable consistent theories.

I wasn't assuming that $T'$ is consistent; the inconsistent theory is the top of the lattice. However, you are right that using Rosser's sentence instead is perfectly justifiable.
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François G. Dorais♦May 30 '10 at 20:51

I don't know if Rosser's sentence gives a theory which is independent of the enumeration of T. Do you happen to know?
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François G. Dorais♦May 30 '10 at 21:30

I doubt it. At the very least, it would seem required for the theory to prove that the enumerations gave the same theory, not merely that this was true.
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Joel David HamkinsMay 30 '10 at 21:45

Not only can Rosser’s sentence depend on the enumeration, but in fact, it can also depend on the method of diagonalization. That is, there exist proof predicates for which Rosser’s fixed point equation has more than one solution up to provable equivalence. This is an old result of (IIRC) Guaspari and Solovay. (In contrast, Gödel’s sentence, and more generally any fixed point equation using just the provability predicate and Boolean connectives, is unique up to provable equivalence, for a fixed proof predicate.)
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Emil JeřábekNov 8 '12 at 11:24